Pipeline Beacon Architect 46 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing pipeline and beacon metrics, and an integer K, construct an optimal algorithm to evaluate and compute the target architect value by summing all values strictly greater than K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Pipeline Beacon Architect 46"
WHY DOES IT MATTER?
The linear scan pattern is essential because it guarantees that the algorithm scales linearly with input size, which is critical for large datasets common in real‑time analytics and monitoring pipelines.
OPTIMIZATION CHALLENGE
The key insight is that the comparison operation is associative and independent, allowing us to avoid any nested loops or auxiliary data structures.
REAL-WORLD CONNECTION
Think of a real‑time monitoring dashboard that aggregates sensor readings above a safety threshold; it must process each reading instantly to trigger alerts, mirroring the O(n) scan.
When explaining this to an interviewer, emphasize that the algorithm’s simplicity is its strength—no extra space, no complex data structures, just a single pass and a running total.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to a simple linear scan over the input sequence, accumulating values that exceed the threshold K. A naive approach might involve nested loops or repeated filtering, leading to O(n^2) time or excessive memory usage. By recognizing that each element is independent of the others for the comparison operation, we can process each element exactly once, achieving O(n) time and O(1) auxiliary space, which is optimal for this class of threshold‑based summation problems.
Interview Questions on This Problem
Q1How would you optimize the solution if the input array is extremely large and cannot fit into memory?
You would stream the data, reading one element at a time, and maintain a running sum of values greater than K. This approach keeps memory usage constant regardless of input size.
Q2In a distributed system, how could you parallelize this computation across multiple nodes?
Partition the array into chunks, assign each chunk to a worker node that computes its local sum of values > K, and then aggregate the partial sums using a reduce operation. This leverages data parallelism while keeping the algorithm linear overall.
Q3What edge cases would you test for in a production environment where K might be negative or the array contains zeros?
Test with K negative (all positive numbers should be summed), K zero (positive numbers only), and arrays containing zeros or negative numbers to ensure they are correctly excluded from the sum.
Examples
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 100], K = 10
Output
100
Explanation: Step-by-step: with input [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 100] and K = 10, we sum all values greater than K. Since only 100 is greater than 10, the output is 100.
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 70], K = 10
Output
70
Explanation: Step-by-step: with input [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 70] and K = 10, we sum all values greater than K. Since only 70 is greater than 10, the output is 70.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
The optimal approach is a single linear scan: for each element, if it exceeds K, add it to a running sum. This runs in O(n) time and uses O(1) extra space.
Brute Force Approach
A naive solution might sort the array and then iterate from the largest element downwards, stopping when elements are no longer greater than K, or use nested loops to compare each element with every other, both of which are unnecessary and inefficient.
Verified Code Solutions
function solution(nums, K) { let sum = 0; for (let num of nums) { if (num > K) { sum += num; } } return sum; }class Solution { public: int solution(vector<int>& nums, int K) { int sum = 0; for (int num : nums) { if (num > K) { sum += num; } } return sum; } };class Solution { public int solution(int[] nums, int K) { int sum = 0; for (int num : nums) { if (num > K) { sum += num; } } return sum; } }def solution(nums, K): return sum(num for num in nums if num > K)function solution(nums, K) { let sum = 0; for (let num of nums) { if (num > K) { sum += num; } } return sum; }Asked in Top Tech Interviews
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